Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
$220$
$210$
$200$
$105$
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
$f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The variance of first $50$ even natural numbers is
The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
Height | Weight | |
Mean | $162.6\,cm$ | $52.36\,kg$ |
Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?